Solve 20=p^2+1+p^2-4p+13

$$20 = p ^ { 2 } + 1 + p ^ { 2 } - 4 p + 13$$

Solve for p

$$p=-1$$

$$p=3$$

Steps Using Factoring By Grouping

Combine p^{2} and p^{2} to get 2p^{2}.

$$20=2p^{2}+1-4p+13$$

Add 1 and 13 to get 14.

$$20=2p^{2}+14-4p$$

Swap sides so that all variable terms are on the left hand side.

$$2p^{2}+14-4p=20$$

Subtract 20 from both sides.

$$2p^{2}+14-4p-20=0$$

Subtract 20 from 14 to get -6.

$$2p^{2}-6-4p=0$$

Divide both sides by 2.

$$p^{2}-3-2p=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$p^{2}-2p-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite p^{2}-2p-3 as \left(p^{2}-3p\right)+\left(p-3\right).

$$\left(p^{2}-3p\right)+\left(p-3\right)$$

Factor out p in p^{2}-3p.

$$p\left(p-3\right)+p-3$$

Factor out common term p-3 by using distributive property.

$$\left(p-3\right)\left(p+1\right)$$

To find equation solutions, solve p-3=0 and p+1=0.

$$p=3 p=-1$$

Steps Using the Quadratic Formula

Combine p^{2} and p^{2} to get 2p^{2}.

$$20=2p^{2}+1-4p+13$$

Add 1 and 13 to get 14.

$$20=2p^{2}+14-4p$$

Swap sides so that all variable terms are on the left hand side.

$$2p^{2}+14-4p=20$$

Subtract 20 from both sides.

$$2p^{2}+14-4p-20=0$$

Subtract 20 from 14 to get -6.

$$2p^{2}-6-4p=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2p^{2}-4p-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square -4.

$$p=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$p=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$p=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

$$p=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$p=\frac{-\left(-4\right)±8}{2\times 2}$$

The opposite of -4 is 4.

$$p=\frac{4±8}{2\times 2}$$

Multiply 2 times 2.

$$p=\frac{4±8}{4}$$

Now solve the equation p=\frac{4±8}{4} when ± is plus. Add 4 to 8.

$$p=\frac{12}{4}$$

Divide 12 by 4.

$$p=3$$

Now solve the equation p=\frac{4±8}{4} when ± is minus. Subtract 8 from 4.

$$p=-\frac{4}{4}$$

Divide -4 by 4.

$$p=-1$$

The equation is now solved.

$$p=3 p=-1$$

Steps for Completing the Square

Combine p^{2} and p^{2} to get 2p^{2}.

$$20=2p^{2}+1-4p+13$$

Add 1 and 13 to get 14.

$$20=2p^{2}+14-4p$$

Swap sides so that all variable terms are on the left hand side.

$$2p^{2}+14-4p=20$$

Subtract 14 from both sides.

$$2p^{2}-4p=20-14$$

Subtract 14 from 20 to get 6.

$$2p^{2}-4p=6$$

Divide both sides by 2.

$$\frac{2p^{2}-4p}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$p^{2}+\left(-\frac{4}{2}\right)p=\frac{6}{2}$$

Divide -4 by 2.

$$p^{2}-2p=\frac{6}{2}$$

Divide 6 by 2.

$$p^{2}-2p=3$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-2p+1=3+1$$

Add 3 to 1.

$$p^{2}-2p+1=4$$

Factor p^{2}-2p+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$p-1=2 p-1=-2$$

Add 1 to both sides of the equation.

$$p=3 p=-1$$